Area of Rectangles, Triangles and Parallelograms
Every question with a fully worked solution.
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Theory
All three areas come from one idea: a base times the perpendicular height, halved for a triangle. Rectangle \(A = lw\); parallelogram \(A = bh\); triangle \(A = \dfrac{1}{2}bh\). A triangle with base \(5\) cm and height \(8\) cm has area \(\dfrac{1}{2} \times 5 \times 8 = 20~\text{cm}^2\). This Year 8 maths page shows how.
Rectangle: \(A = l \times w\). Parallelogram: \(A = b \times h\). Triangle: \(A = \dfrac{1}{2} \times b \times h\), half of the parallelogram on the same base and height.
The perpendicular height is the distance from the base to the opposite side or vertex, meeting the base at a right angle. A sloping side is never the height.
Any side can be the base, as long as the height is measured to it. When the shape leans, the height falls outside, down to the base line made longer; the extension is not part of the base.
On a number plane, take the base along a grid line and count grid units for the base and the height.
Reverse problems. Substitute what you know, then solve. For a triangle, double both sides first to remove the half.
Units. Put the lengths in one unit before multiplying; the answer is in square units. Composite shapes split into these three: add the parts, or take a piece away.
Base times perpendicular height:
| Shape | Area |
|---|---|
| rectangle | \(A = l \times w\) |
| parallelogram | \(A = b \times h\) |
| triangle | \(A = \dfrac{1}{2} \times b \times h\) |
How to find an area
- Choose the formula and identify the base and the perpendicular height.
- Change the lengths to one unit.
- Substitute, calculate and give square units. For a reverse problem, solve for the missing length.
(ii) Find the area of a triangle with height \(8\) cm and base \(5\) cm.
| \(A\) | \(=\) | \(3 \times 5 = 15\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 5 \times 8 = 20\) |
(i) \(15~\text{cm}^2\)
(ii) \(20~\text{cm}^2\)
(ii) A parallelogram has an area of \(48~\text{cm}^2\) and a height of \(6\) cm. Find its base.
| \(A\) | \(=\) | \(6 \times 7 = 42\) |
| \(b \times 6\) | \(=\) | \(48\) |
| \(b\) | \(=\) | \(48 \div 6 = 8\) |
(i) \(42~\text{cm}^2\)
(ii) \(8\) cm
The \(9.5\) cm side slopes. Only the dashed \(7\) cm line meets the base at a right angle.
| \(A\) | \(=\) | \(b \times h\) |
| \(\) | \(=\) | \(12 \times 7 = 84\) |
\(84~\text{cm}^2\)
Change the base to metres: \(240 \div 100 = 2.4\) m.
| \(1.8\) | \(=\) | \(\dfrac{1}{2} \times 2.4 \times h\) |
| \(1.8\) | \(=\) | \(1.2 \times h\) |
| \(h\) | \(=\) | \(1.8 \div 1.2 = 1.5\) |
The height is \(1.5\) m. Check: \(\dfrac{1}{2} \times 2.4 \times 1.5 = 1.8\).
| \(\text{banner}\) | \(=\) | \(2.4 \times 1.5 = 3.6\) |
| \(\text{logo}\) | \(=\) | \(0.8 \times 0.45 = 0.36\) |
| \(3.6 - 0.36\) | \(=\) | \(3.24\) |
\(3.24~\text{m}^2\) of banner is left.
(i) Its base is \(8\) cm. Find its height.
(ii) A triangle with the same area has a height of \(5\) cm. Find its base.
(iii) The parallelogram’s base is doubled and its height halved. Find the new area.
| \(h\) | \(=\) | \(60 \div 8 = 7.5\) |
| \(\dfrac{1}{2} \times b \times 5\) | \(=\) | \(60\) |
| \(b\) | \(=\) | \(120 \div 5 = 24\) |
| \(16 \times 3.75\) | \(=\) | \(60\) |
(i) \(7.5\) cm
(ii) \(24\) cm
(iii) \(60~\text{cm}^2\): doubling and halving cancel out.
Common pitfalls
Frequently asked questions
What is the area of a parallelogram?
Base times perpendicular height: \(A = bh\).
Why is a triangle half of bh?
A diagonal cuts a parallelogram into two identical triangles.
What is the perpendicular height?
The distance from the base to the opposite side or vertex, at right angles to the base.
How do I find the base from the area?
For a parallelogram, \(b = A \div h\); for a triangle, \(b = 2A \div h\).
What units is area in?
Square units, such as \(\text{cm}^2\), \(\text{m}^2\) or hectares.